a) x² - 13x = 0
⇔ x.(x - 13) = 0
⇔ \(\left[ \begin{array}{l}x = 0\\x - 13 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 0\\x = 13\end{array} \right.\)
Vậy x ∈ {0; 13}
b) 2x.(x - 2) + 3.(x - 2) = 0
⇔ (x - 2)(2x + 3) = 0
⇔ \(\left[ \begin{array}{l}x - 2 = 0\\2x + 3 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 2\\x = -\frac{3}{2}\end{array} \right.\)
Vậy x ∈ {2; -$\frac{3}{2}$}
c) x + 1 = (x + 1)²
⇔ (x +1) - (x + 1)² = 0
⇔ (x + 1)(1 - x - 1) = 0
⇔ (x + 1).(-x) = 0
⇔ \(\left[ \begin{array}{l}x + 1 = 0\\-x = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = -1\\x = 0\end{array} \right.\)
Vậy x ∈ {-1; 0}
d) x.(x - 2) + x - 2 = 0
⇔ x(x - 2) + (x - 2) = 0
⇔ (x - 2)(x + 1) = 0
⇔ \(\left[ \begin{array}{l}x - 2 = 0\\x + 1 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 2\\x = -1\end{array} \right.\)
Vậy x ∈ {2; -1}
